Find the smallest positive real number $x$ such that
\[\lfloor x^2 \rfloor - x \lfloor x \rfloor = 6.\]
Solution: Let $n = \lfloor x \rfloor$ and $f = \{x\}.$  Then $x = n + f,$ so
\[\lfloor n^2 + 2nf + f^2 \rfloor - (n + f) n = 6.\]Since $n^2$ is an integer, we can pull it out of the floor, to get
\[n^2 + \lfloor 2nf + f^2 \rfloor - n^2 - nf = 6.\]Thus,
\[\lfloor 2nf + f^2 \rfloor - nf = 6.\]Since $\lfloor 2nf + f^2 \rfloor$ and 6 are integers, $nf$ must also be an integer.  Hence, we can also pull $2nf$ out of the floor, to get
\[2nf + \lfloor f^2 \rfloor = nf + 6,\]so $nf + \lfloor f^2 \rfloor = 6.$

Since $0 \le f < 1,$ $0 \le f^2 < 1,$ so $\lfloor f^2 \rfloor = 0.$  Hence, $nf = 6,$ so
\[n = \frac{6}{f}.\]Since $f < 1,$ $n > 6.$  The smallest possible value of $n$ is then 7.  If $n = 7,$ then $f = \frac{6}{7},$ so $x = 7 + \frac{6}{7} = \frac{55}{7},$ which is a solution.  Thus, the smallest solution $x$ is $\boxed{\frac{55}{7}}.$